an object is thrown upward with an initial velocity 'u' m/s. derive an expression for the height 'h' at which its kinetic energy becomes equal to its potential energy.

Answer 1

h=(u²- v²)/2g
v²=u²- 2gh,
KE=m•v²/2= m(u²- 2gh)/2,
KE=PE =>
m(u²- 2gh)/2 =mgh,
(u²- 2gh)/2= gh,
u²=4gh,
h= u²/4g.